Question

{"given":"Doubly ionized lithium atom: $Z = 3$, ground state $n = 1$. Energy formula for hydrogen-like atoms: $$E_n = -13.6 \\frac{Z^2}{n^2} \\text{ eV}$$","key_observation":"For a hydrogen-like atom with atomic number $Z$, the ground state energy scales as $Z^2$, so the ionization energy from the ground state is $13.6 \\times Z^2$ eV.","option_analysis":[{"label":"(A)","text":"1.51 eV","verdict":"incorrect","explanation":"$1.51 \\text{ eV}$ corresponds to the ionization energy from the $n=3$ level of hydrogen ($13.6/9$), not the ground state of Li$^{2+}$."},{"label":"(B)","text":"13.6 eV","verdict":"incorrect","explanation":"$13.6 \\text{ eV}$ is the ionization energy of hydrogen ($Z=1, n=1$), not doubly ionized lithium."},{"label":"(C)","text":"40.8 eV","verdict":"incorrect","explanation":"$40.8 \\text{ eV} = 13.6 \\times 3$ corresponds to incorrect use of $Z$ instead of $Z^2$ in the formula."},{"label":"(D)","text":"122.4 eV","verdict":"correct","explanation":"$$E_1 = -13.6 \\times \\frac{(3)^2}{(1)^2} = -13.6 \\times 9 = -122.4 \\text{ eV}$$ So the ionization energy is $122.4 \\text{ eV}$."}],"answer":"(D)","formula_steps":[]}